[Lintcode] Continuous Subarray Sum

Given an integer array, find a continuous subarray where the sum of numbers is the biggest. Your code should return the index of the first number and the index of the last number. (If their are duplicate answer, return anyone)

Example

Give [-3, 1, 3, -3, 4], return [1,4].

找最大连续子集和并返回区间坐标。解法是 遍历数组元素，存储max，当当前sum大于max时，储存区间信息。当sum小于0时，初始化区间信息，重新开始计算sum。 因为重新计算的话可以将之前的sum看作0，肯定会比之前小于0的sum值要大。遍历同时储存最大元素及其位置，如果全部为负数的话，返回该位置信息。

public class Solution {    /**     * @param A an integer array     * @return  A list of integers includes the index of the first number and the index of the last number     */    public ArrayList<Integer> continuousSubarraySum(int[] A) {        ArrayList<Integer> res = new ArrayList<Integer>();        if(A.length == 0) return res;        //if sum < 0, start in new position        int start = 0;        int max = Integer.MIN_VALUE;        int sum = 0;        int maxNeg = Integer.MIN_VALUE, minIndex = 0;        for(int i = 0; i < A.length; i++) {            if(A[i] > maxNeg) {maxNeg = A[i]; minIndex = i;}            sum += A[i];            if(sum <= 0) {sum = 0; start = i + 1;}            else {                if(sum > max) {                    max = sum;                    res.clear();                    res.add(start);                    res.add(i);                }            }        }                if(max < 0) {            res.add(minIndex);            res.add(minIndex);            return res;        }        return res;    }}